details

property	value
status	complete
benchmark	poly4.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n020.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	1.07128000259 seconds
cpu usage	1.098618673
max memory	1.2591104E7

stage attributes

key	value
output-size	4914
starexec-result	YES

output

/export/starexec/sandbox/solver/bin/starexec_run_Itrs /export/starexec/sandbox/benchmark/theBenchmark.itrs /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES DP problem for innermost termination. P = eval#(an, bn, cn, i, j) -> eval#(an, bn, cn + 1, i, j + 1) [j < bn && i >= an] eval#(I0, I1, I2, I3, I4) -> eval#(I0, I1, I2 + 1, I3 + 1, I4) [I4 >= I1 && I3 < I0] eval#(I5, I6, I7, I8, I9) -> eval#(I5, I6, I7 + 1, I8 + 1, I9) [I9 < I6 && I8 < I5] eval#(I10, I11, I12, I13, I14) -> eval#(I10, I11, I12 + 1, I13, I14 + 1) [I14 < I11 && I13 < I10] R = eval(an, bn, cn, i, j) -> eval(an, bn, cn + 1, i, j + 1) [j < bn && i >= an] eval(I0, I1, I2, I3, I4) -> eval(I0, I1, I2 + 1, I3 + 1, I4) [I4 >= I1 && I3 < I0] eval(I5, I6, I7, I8, I9) -> eval(I5, I6, I7 + 1, I8 + 1, I9) [I9 < I6 && I8 < I5] eval(I10, I11, I12, I13, I14) -> eval(I10, I11, I12 + 1, I13, I14 + 1) [I14 < I11 && I13 < I10] The dependency graph for this problem is: 0 -> 0 1 -> 1 2 -> 0, 2, 3 3 -> 1, 2, 3 Where: 0) eval#(an, bn, cn, i, j) -> eval#(an, bn, cn + 1, i, j + 1) [j < bn && i >= an] 1) eval#(I0, I1, I2, I3, I4) -> eval#(I0, I1, I2 + 1, I3 + 1, I4) [I4 >= I1 && I3 < I0] 2) eval#(I5, I6, I7, I8, I9) -> eval#(I5, I6, I7 + 1, I8 + 1, I9) [I9 < I6 && I8 < I5] 3) eval#(I10, I11, I12, I13, I14) -> eval#(I10, I11, I12 + 1, I13, I14 + 1) [I14 < I11 && I13 < I10] We have the following SCCs. { 2, 3 } { 0 } { 1 } DP problem for innermost termination. P = eval#(I0, I1, I2, I3, I4) -> eval#(I0, I1, I2 + 1, I3 + 1, I4) [I4 >= I1 && I3 < I0] R = eval(an, bn, cn, i, j) -> eval(an, bn, cn + 1, i, j + 1) [j < bn && i >= an] eval(I0, I1, I2, I3, I4) -> eval(I0, I1, I2 + 1, I3 + 1, I4) [I4 >= I1 && I3 < I0] eval(I5, I6, I7, I8, I9) -> eval(I5, I6, I7 + 1, I8 + 1, I9) [I9 < I6 && I8 < I5] eval(I10, I11, I12, I13, I14) -> eval(I10, I11, I12 + 1, I13, I14 + 1) [I14 < I11 && I13 < I10] We use the reverse value criterion with the projection function NU: NU[eval#(z1,z2,z3,z4,z5)] = z1 + -1 * z4 This gives the following inequalities: I4 >= I1 && I3 < I0 ==> I0 + -1 * I3 > I0 + -1 * (I3 + 1) with I0 + -1 * I3 >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. DP problem for innermost termination. P = eval#(an, bn, cn, i, j) -> eval#(an, bn, cn + 1, i, j + 1) [j < bn && i >= an] R = eval(an, bn, cn, i, j) -> eval(an, bn, cn + 1, i, j + 1) [j < bn && i >= an] eval(I0, I1, I2, I3, I4) -> eval(I0, I1, I2 + 1, I3 + 1, I4) [I4 >= I1 && I3 < I0] eval(I5, I6, I7, I8, I9) -> eval(I5, I6, I7 + 1, I8 + 1, I9) [I9 < I6 && I8 < I5] eval(I10, I11, I12, I13, I14) -> eval(I10, I11, I12 + 1, I13, I14 + 1) [I14 < I11 && I13 < I10] We use the reverse value criterion with the projection function NU: NU[eval#(z1,z2,z3,z4,z5)] = z2 + -1 * z5 This gives the following inequalities: j < bn && i >= an ==> bn + -1 * j > bn + -1 * (j + 1) with bn + -1 * j >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. DP problem for innermost termination. P = eval#(I5, I6, I7, I8, I9) -> eval#(I5, I6, I7 + 1, I8 + 1, I9) [I9 < I6 && I8 < I5] eval#(I10, I11, I12, I13, I14) -> eval#(I10, I11, I12 + 1, I13, I14 + 1) [I14 < I11 && I13 < I10] R = eval(an, bn, cn, i, j) -> eval(an, bn, cn + 1, i, j + 1) [j < bn && i >= an] eval(I0, I1, I2, I3, I4) -> eval(I0, I1, I2 + 1, I3 + 1, I4) [I4 >= I1 && I3 < I0] eval(I5, I6, I7, I8, I9) -> eval(I5, I6, I7 + 1, I8 + 1, I9) [I9 < I6 && I8 < I5] eval(I10, I11, I12, I13, I14) -> eval(I10, I11, I12 + 1, I13, I14 + 1) [I14 < I11 && I13 < I10] We use the reverse value criterion with the projection function NU: NU[eval#(z1,z2,z3,z4,z5)] = z1 + -1 * z4 This gives the following inequalities: I9 < I6 && I8 < I5 ==> I5 + -1 * I8 > I5 + -1 * (I8 + 1) with I5 + -1 * I8 >= 0 I14 < I11 && I13 < I10 ==> I10 + -1 * I13 >= I10 + -1 * I13 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = eval#(I10, I11, I12, I13, I14) -> eval#(I10, I11, I12 + 1, I13, I14 + 1) [I14 < I11 && I13 < I10] R = eval(an, bn, cn, i, j) -> eval(an, bn, cn + 1, i, j + 1) [j < bn && i >= an] eval(I0, I1, I2, I3, I4) -> eval(I0, I1, I2 + 1, I3 + 1, I4) [I4 >= I1 && I3 < I0] eval(I5, I6, I7, I8, I9) -> eval(I5, I6, I7 + 1, I8 + 1, I9) [I9 < I6 && I8 < I5] eval(I10, I11, I12, I13, I14) -> eval(I10, I11, I12 + 1, I13, I14 + 1) [I14 < I11 && I13 < I10] We use the reverse value criterion with the projection function NU: NU[eval#(z1,z2,z3,z4,z5)] = z2 + -1 * z5 popout

output may be truncated. 'popout' for the full output.

job log

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actions

all output return to Integ TRS Inner 43557